I have the following givens:
Let $f$ be a nonnegative function on a measure space $(X, \mathfrak{M}, \mu)$ and suppose that $f$ is integrable with respect to $\mu$. for each $E \in \mathfrak{M}$ define $$\nu(E) = \int_{E} f d\mu.$$
And I am trying to prove that $\nu$ is an extended real-valued function from $\mathfrak{M}$ to $[0,\infty].$ but I know that the definition of the integral of a nonnegative function $f$ differs from the case of Lebesgue integral as we are in a general measure space. my question is:
Is $\int_{E} f d\mu $ still equals to $f \mu (E)$ as incase of Lebesgue measure?If so why? because this will prove that $\nu$ is an extended real-valued function from $\mathfrak{M}$ to $[0,\infty],$ which is what I want.
Suppose $\nu(E) \not\in [0,+\infty]$. i.e: $\nu(E)<0$. So, $\int_E fd \mu = sup \int_E s d\mu<0$, the supremum is taken over all simple functions $0<s \leq f$.
By definition of the integral of unsigned simple functions, $\int_Es d\mu = \sum_{i=1}^{n}c_i \mu(E_i)$, where $c_i \in [0, \infty]$ and {$E_i$} is a measurable partition of $E$.
What can we say about $\int_Es d\mu$?. This is clearly non-negative since it's a finite sum of non-negative terms ($c_i \geq 0$ & $\mu(E_i) \geq 0$since $\mu$ is a measure).
Therefore, sup $\int_E s d\mu<0$ for all $0<s \leq f$ , is a contradiction. $\nu(E)\geq 0$